I'm getting two different answers for the following series...
$$\sum_{k=0}^{\infty} q^{2k}$$
With $-1 < q < 1$, I can solve the series by using the infinite geometric sum formula.
$$\sum_{k=0}^{\infty} q^{2k} = \frac{1}{1-q^2}$$
However, if I make the substitution, $m=2k$, then I get a different result.
$$\sum_{k=0}^{\infty} q^{2k} = \sum_{m=0}^{\infty} q^{m} = \frac{1}{1-q}$$
Where am I going wrong?
On
What you were doing with the substitution $m = 2k$ was resulting in $$\sum_{m=0}^∞ q^m = q^0+q^1+q^2+q^3+q^4...$$ but the original equation yields $$\sum_{k=0}^∞ q^{2k} = q^0+q^2+q^4...$$ leaving a difference of $$\sum_{k=0}^∞ q^{2k+1} = q^1+q^3+q^5...$$ Instead of $m = 2k$, have $m = q^2$. Then, from $-1 < q < 1$ we get $0 \leq m < 1$. Then, $$\sum_{k=0}^∞ (q^2)^k = \sum_{k=0}^∞ (m)^k$$ and from this, because of the domain of m, we can apply the sum of an infinite geometric series formula. $$\frac{1}{1-m}=\frac{1}{1-q^2}$$
By making the substitution $m=2k$ you are creating a sum that adds all the powers of $q$, not just the even powers, which is what your original sum does. For example, $q^3$ appears in the right-hand sum, but not in the original sum.